
## 7.49 Sheafification in a topology

In this section we explain the analogue of the sheafification construction in a topology.

Let $\mathcal{C}$ be a category. Let $J$ be a topology on $\mathcal{C}$. Let $\mathcal{F}$ be a presheaf of sets. For every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ we define

$L\mathcal{F}(U) = \mathop{\mathrm{colim}}\nolimits _{S \in J(U)^{opp}} \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, \mathcal{F})$

as a colimit. Here we think of $J(U)$ as a partially ordered set, ordered by inclusion, see Lemma 7.47.2. The transition maps in the system are defined as follows. If $S \subset S'$ are in $J(U)$, then $S \to S'$ is a morphism of presheaves. Hence there is a natural restriction mapping

$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, \mathcal{F}) \longrightarrow \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S', \mathcal{F}).$

Thus we see that $S \mapsto \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, \mathcal{F})$ is a directed system as in Categories, Definition 4.21.2 provided we reverse the ordering on $J(U)$ (which is what the superscript ${}^{opp}$ is supposed to indicate). In particular, since $h_ U \in J(U)$ there is a canonical map

$\ell : \mathcal{F}(U) \longrightarrow L\mathcal{F}(U)$

coming from the identification $\mathcal{F}(U) = \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ U, \mathcal{F})$. In addition, the colimit defining $L\mathcal{F}(U)$ is directed since for any pair of covering sieves $S, S'$ on $U$ the sieve $S \cap S'$ is a covering sieve too, see Lemma 7.47.2.

Let $f : V \to U$ be a morphism in $\mathcal{C}$. Let $S \in J(U)$. There is a commutative diagram

$\xymatrix{ S \times _ U V \ar[r] \ar[d] & h_ V \ar[d] \\ S \ar[r] & h_ U }$

We can use the left vertical map to get canonical restriction maps

$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, \mathcal{F}) \to \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S \times _ U V, \mathcal{F}).$

Base change $S \mapsto S \times _ U V$ induces an order preserving map $J(U) \to J(V)$. And the restriction maps define a transformation of functors as in Categories, Lemma categories-lemma-functorial-colimit. Hence we get a natural restriction map

$L\mathcal{F}(U) \longrightarrow L\mathcal{F}(V).$

Lemma 7.49.1. In the situation above.

1. The assignment $U \mapsto L\mathcal{F}(U)$ combined with the restriction mappings defined above is a presheaf.

2. The maps $\ell$ glue to give a morphism of presheaves $\ell : \mathcal{F} \to L\mathcal{F}$.

3. The rule $\mathcal{F} \mapsto (\mathcal{F} \xrightarrow {\ell } L\mathcal{F})$ is a functor.

4. If $\mathcal{F}$ is a subpresheaf of $\mathcal{G}$, then $L\mathcal{F}$ is a subpresheaf of $L\mathcal{G}$.

5. The map $\ell : \mathcal{F} \to L\mathcal{F}$ has the following property: For every section $s \in L\mathcal{F}(U)$ there exists a covering sieve $S$ on $U$ and an element $\varphi \in \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, \mathcal{F})$ such that $\ell (\varphi )$ equals the restriction of $s$ to $S$.

Proof. Omitted. $\square$

Definition 7.49.2. Let $\mathcal{C}$ be a category. Let $J$ be a topology on $\mathcal{C}$. We say that a presheaf of sets $\mathcal{F}$ is separated if for every object $U$ and every covering sieve $S$ on $U$ the canonical map $\mathcal{F}(U) \to \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, \mathcal{F})$ is injective.

Theorem 7.49.3. Let $\mathcal{C}$ be a category. Let $J$ be a topology on $\mathcal{C}$. Let $\mathcal{F}$ be a presheaf of sets.

1. The presheaf $L\mathcal{F}$ is separated.

2. If $\mathcal{F}$ is separated, then $L\mathcal{F}$ is a sheaf and the map of presheaves $\mathcal{F} \to L\mathcal{F}$ is injective.

3. If $\mathcal{F}$ is a sheaf, then $\mathcal{F} \to L\mathcal{F}$ is an isomorphism.

4. The presheaf $LL\mathcal{F}$ is always a sheaf.

Proof. Part (3) is trivial from the definition of $L$ and the definition of a sheaf (Definition 7.47.10). Part (4) follows formally from the others.

We sketch the proof of (1). Suppose $S$ is a covering sieve of the object $U$. Suppose that $\varphi _ i \in L\mathcal{F}(U)$, $i = 1, 2$ map to the same element in $\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S, L\mathcal{F})$. We may find a single covering sieve $S'$ on $U$ such that both $\varphi _ i$ are represented by elements $\varphi _ i \in \mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S', \mathcal{F})$. We may assume that $S' = S$ by replacing both $S$ and $S'$ by $S' \cap S$ which is also a covering sieve, see Lemma 7.47.2. Suppose $V\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and $\alpha : V \to U$ in $S(V)$. Then we have $S \times _ U V = h_ V$, see Lemma 7.47.5. Thus the restrictions of $\varphi _ i$ via $V \to U$ correspond to sections $s_{i, V, \alpha }$ of $\mathcal{F}$ over $V$. The assumption is that there exist a covering sieve $S_{V, \alpha }$ of $V$ such that $s_{i, V, \alpha }$ restrict to the same element of $\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S_{V, \alpha }, \mathcal{F})$. Consider the sieve $S''$ on $U$ defined by the rule

7.49.3.1
\begin{eqnarray} \label{sites-equation-S-prime-prime} (f : T \to U) \in S''(T) & \Leftrightarrow & \exists \ V , \ \alpha : V \to U, \ \alpha \in S(V), \nonumber \\ & & \exists \ g : T \to V, \ g \in S_{V, \alpha }(T), \\ & & f = \alpha \circ g \nonumber \end{eqnarray}

By axiom (2) of a topology we see that $S''$ is a covering sieve on $U$. By construction we see that $\varphi _1$ and $\varphi _2$ restrict to the same element of $\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{C})}(S'', L\mathcal{F})$ as desired.

We sketch the proof of (2). Assume that $\mathcal{F}$ is a separated presheaf of sets on $\mathcal{C}$ with respect to the topology $J$. Let $S$ be a covering sieve of the object $U$ of $\mathcal{C}$. Suppose that $\varphi \in \mathop{Mor}\nolimits _\mathcal {C}(S, L\mathcal{F})$. We have to find an element $s \in L\mathcal{F}(U)$ restricting to $\varphi$. Suppose $V\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, and $\alpha : V \to U$ in $S(V)$. The value $\varphi (\alpha ) \in L\mathcal{F}(V)$ is given by a covering sieve $S_{V, \alpha }$ of $V$ and a morphism of presheaves $\varphi _{V, \alpha } : S_{V, \alpha } \to \mathcal{F}$. As in the proof above, define a covering sieve $S''$ on $U$ by Equation (7.49.3.1). We define

$\varphi '' : S'' \longrightarrow \mathcal{F}$

by the following simple rule: For every $f : T \to U$, $f \in S''(T)$ choose $V, \alpha , g$ as in Equation (7.49.3.1). Then set

$\varphi ''(f) = \varphi _{V, \alpha }(g).$

We claim this is independent of the choice of $V, \alpha , g$. Consider a second such choice$V', \alpha ', g'$. The restrictions of $\varphi _{V, \alpha }$ and $\varphi _{V', \alpha '}$ to the intersection of the following covering sieves on $T$

$(S_{V, \alpha } \times _{V, g} T) \cap (S_{V', \alpha '} \times _{V', g'} T)$

agree. Namely, these restrictions both correspond to the restriction of $\varphi$ to $T$ (via $f$) and the desired equality follows because $\mathcal{F}$ is separated. Denote the common restriction $\psi$. The independence of choice follows because $\varphi _{V, \alpha }(g) = \psi (\text{id}_ T) = \varphi _{V', \alpha '}(g')$. OK, so now $\varphi ''$ gives an element $s \in L\mathcal{F}(U)$. We leave it to the reader to check that $s$ restricts to $\varphi$. $\square$

Definition 7.49.4. Let $\mathcal{C}$ be a category endowed with a topology $J$. Let $\mathcal{F}$ be a presheaf of sets on $\mathcal{C}$. The sheaf $\mathcal{F}^\# := LL\mathcal{F}$ together with the canonical map $\mathcal{F} \to \mathcal{F}^\#$ is called the sheaf associated to $\mathcal{F}$.

Proposition 7.49.5. Let $\mathcal{C}$ be a category endowed with a topology. Let $\mathcal{F}$ be a presheaf of sets on $\mathcal{C}$. The canonical map $\mathcal{F} \to \mathcal{F}^\#$ has the following universal property: For any map $\mathcal{F} \to \mathcal{G}$, where $\mathcal{G}$ is a sheaf of sets, there is a unique map $\mathcal{F}^\# \to \mathcal{G}$ such that $\mathcal{F} \to \mathcal{F}^\# \to \mathcal{G}$ equals the given map.

Proof. Same as the proof of Proposition 7.10.12. $\square$

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